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HAL Id: hal-01491850

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A staggered scheme for the Euler equations

Thierry Goudon, Julie Llobell, Sebastian Minjeaud

To cite this version:

Thierry Goudon, Julie Llobell, Sebastian Minjeaud. A staggered scheme for the Euler equations.

Finite Volumes for Complex Applications VIII, 2017, Lille, France. pp.91-99. �hal-01491850�

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A staggered scheme for the Euler equations.

Thierry Goudon, Julie Llobell and Sebastian Minjeaud

Abstract We extend to the full Euler system the scheme introduced in [Berthelin, Goudon, Minjeaud, Math. Comp. 2014] for solving the barotropic Euler equations. This finite volume scheme is defined on staggered grids with numerical fluxes de-rived in the spirit of kinetic schemes. The difficulty consists in finding a suitable treatment of the energy equation while density and internal energy on the one hand, and velocity on the other hand, are naturally defined on dual locations. The proposed scheme uses the density, the velocity and the internal energy as computational vari-ables and stability conditions are identified in order to preserve the positivity of the discrete density and internal energy. Moreover, we define averaged energies which satisfy local conservation equations. Finally, we provide numerical simulations of Riemann problems to illustrate the behaviour of the scheme.

Key words: Finite Volumes, Conservation laws, Staggered grids, Euler equations MSC (2010): 65M08, 76M12, 35L65, 35Q31

1 Introduction.

This work aims at designing a scheme to numerically solve the 1D-Euler system:

∂t   ρ ρ u ρ E  + ∂x   ρ u ρ u2+ p ρ Eu + pu  = 0, (t, x) ∈ [0, ∞) × R, E= u2/2 + e, p= (γ − 1)ρe, (1)

Thierry Goudon, e-mail: thierry.goudon@inria.fr Julie Llobell, e-mail: llobell@unice.fr

Sebastian Minjeaud, e-mail: sebastian.minjeaud@unice.fr Universit´e Cˆote d’Azur, CNRS, Inria, LJAD, France

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2 Thierry Goudon, Julie Llobell and Sebastian Minjeaud

where ρ, u, E and p stand for the density, the velocity, the total energy and the pressure respectively, u2/2 and e are the kinetic and internal energies, and γ > 1 is the adiabatic exponent.

We wish to extend to (1) the scheme designed in [1] for the barotropic Euler equations. This scheme works on staggered grids – meaning that densities and ve-locities are not collocated – and this raises a difficulty for (1) as the definition of the total energy mixes quantities, namely the velocity and the internal energy, naturally defined on different grids. To address this issue, it is convenient to work with the internal energy equation, namely

∂t(ρe) + ∂x(ρeu) = −p∂xu, (2) instead of the evolution equation for ρE, since discrete densities, pressures, and in-ternal energies are naturally stored at the same locations. This formulation has also the advantage of making more direct the analysis of the positivity of e. Unfortu-nately, as it is well-known, this non conservative formulation is not equivalent to (1) when the solution presents discontinuities. We shall follow the approach discussed in [2] by introducing in (2) correction terms accounting for the kinetic energy bal-ance. Then, the scheme introduced in [2] can be shown: a) to be consistent with (a weak form of) the total energy equation as the space step δ x goes to zero and b) to conserve the global discrete total energy. Our purpose is two-fold. First of all, we shall adapt the scheme of [1] for dealing with (1). Second of all, we introduce aver-aged energies which satisfy local conservation equations. Finally we provide some numerical simulations in Section 5.

2 Staggered scheme.

Let (xj)j be a subdivision of the 1D computational domain and denote the size of the cells by δ xj+1

2

= xj+1− xj. The cell centers, xj+1 2

= (xj+ xj+1)/2, define the dual mesh and we set δ xj= (δ xj−1

2 + δ xj+ 1

2)/2. The discrete densities ρj+ 1 2

and internal energies ej+1

2 are stored at the centers xj+12 whereas the velocities uj are

located at the edges xj. The time dicretization is explicit and we use the convention that, with q the evaluation of a certain quantity at time t, q stands for its update at time t + δ t.

• The density ρj+1

2 is updated using the discrete mass balance equation:

ρj+1 2− ρj+12 δ t +Fj+1−Fj δ xj+1 2 = 0.

The mass fluxes are defined byFj=Fj++F − j whereF + j =F+(ρj−1 2, ej, uj), F− j =F −

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the sound speed, the definition of the numerical fluxesF±is extracted from [1] F+(ρ, e, u) =        0 if u6 −c(e), ρ (u + c(e))2

4c(e) if |u| < c(e), ρ u if u > c(e),

andF−(ρ, e, u) = −F+(ρ, e, −u).

In the sequel we use the following two properties: ∀u ∈ R, ∀ρ, e > 0,

06 F+(ρ, e, u) 6 ρ[λ+(e, u)]+ and − ρ[λ−(e, u)]−6 F−(ρ, e, u) 6 0, (3) where λ±(e, u) = u ± c(e) and [z]±=12(|z| ± z).

• The velocity ujis then updated using the discrete momentum balance equation: ρjuj− ρjuj δ t + Gj+1 2−Gj−12 δ xj + Πj+1 2 − Πj−12 δ xj = 0. (4)

The momentum fluxGj+1

2 and the pressure Πj+12 are defined by:

Gj+1 2 = ujF + j+12+ uj+1F − j+12 and Πj+12 = (γ − 1)ρj+12ej+12.

The quantities ρjandF±j+1 2

are expressed as mean values of ρ1 2 andF ± j ,F ± j+1: ρj= δ xj+1 2ρj+12+ δ xj−12ρj−12 2δ xj and F± j+12 = F± j+1+F ± j 2 . (5)

• The internal energy ej+1

2 is updated using the following discrete equation:

ρj+1 2ej+ 1 2− ρj+ 1 2ej+ 1 2 δ t + Ej+1−Ej δ xj+1 2 + Πj+1 2 uj+1− uj δ xj+1 2 = Sj+1 2 . (6)

The internal energy fluxEjis given byEj= ej−1 2F + j + ej+1 2F − j . According to [2], the rhs Sj+1

2 is designed to account for the rest term that appears in the discrete

kinetic energy balance and that do not vanish when δ x goes to zero.

• To be more specific, the kinetic energy balance is obtained by multiplying (4) by uj. We find, see [1, 2]: ρj u2j 2 − ρj u2j 2 δ t + Hj+1 2−Hj−12 δ xj + Πj+1 2− Πj−12 δ xj uj= −Rj,

where the kinetic energy flux is given byHj+1 2 = u2j 2F + j+12+ u2 j+1 2 F − j+12 and

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4 Thierry Goudon, Julie Llobell and Sebastian Minjeaud Rj= 1 2δ tρj(uj− uj) 2+ 1 δ xj  (uj− uj−1)2 2 F + j−1 2 −(uj+1− uj) 2 2 F − j+12  + 1 δ xj (uj− uj)(uj− uj−1)Fj−+1 2 + 1 δ xj (uj− uj)(uj+1− uj)F−j+1 2 .

It is thus quite natural to define the source term in the following way:

Sj+1 2 =δ xj+1Rj+1+ δ xjRj 2δ xj+1 2 .

The scheme presented above is close to the 1D version of the scheme presented in [3] but the two schemes differ by two points. Firstly, the mass fluxes in [3] are upwinded with respect to the material velocity (in other words, it corresponds to the choiceF±(ρ, e, u) = ±ρ[u]±). Secondly, the time steppings are different: even if both schemes are explicit, the variables are not updated in the same order. We solve the discrete equations in the following way: ρ → u → e whereas [3] proceeds with ρ → e → u. In particular, here the corrective term Sj+1

2 does not need any time shift

since the updated velocity u is known when solving (6).

3 Stability conditions.

We now turn to the study of the stability conditions which ensure the positivity of the density and the internal energy.

Proposition 1. Assuming that ej+1

2 > 0, ρj+ 1

2 > 0, ∀ j and the following CFL-like

conditions hold for all j

δ t δ xj+1 2 [uj+1]++ c(ej+3 2) + c(e j+12) 2 + [uj] −+c(ej+12) + c(e j−12) 2 ! 61 γ, (7) δ t δ xj+1 2 c(ej+1 2+k) 6 (γ − 1) 2√2 , ∀k ∈ {−1, 0, 1}, (8) then ej+1 2 > 0 and ρj+ 1 2 > 0.

Proof. We assume that ej+1

2 > 0, ρj+ 1

2 > 0 and that (7) and (8) holds for all j.

We start by observing that:

[λ±(e, u)]±6 [u]±+ c(e) (9) √

2 c(ej) 6 c(ej−1

2) + c(ej+12) (10)

Positivity of the density:As proved in [1], the positivity of ρj+1

2 comes from the

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δ t δ xj+1

2

[λ+(ej+1, uj+1)]++ [λ−(ej, uj)]− 6 1.

It is directly implied by (7) since γ > 1 and (9) holds.

Positivity of the internal energy:We rewrite the terms (−1)iΠj+1

2uj+i, i ∈ {0, 1},

involved in (6), by making the discrete time derivative (uj+i− uj+i) appear. Then, we make use of the Young inequality as follows:

(−1)iΠj+1 2uj+i= (−1) i(γ − 1) ρj+1 2ej+ 1 2 (uj+i− uj+i) + ρj+1 2ej+ 1 2uj+i  > −ρj+12 c(ej+1 2 ) 2√2γ (uj+i− uj+i) 2+ (γ − 1)e j+12 c(ej+1 2 ) √ 2 − (−1) iu j+i !! . Next, we write ρj+1 2ej+12 > T0+ T 0 1+ T11where: T0= ρj+1 2ej+12 1 − δ t δ xj+1 2 (γ − 1) 2 c(ej+1 2) √ 2 − uj+ uj+1 !! − δtEj+1−Ej δ xj+1 2 , Ti1=δ t 2 δ xj+i δ xj+1 2 Rj+i− δ t δ xj+1 2 c(ej+1 2 ) 2√2γ ρj+12 (uj+i− uj+i)2.

Thus, to guarantee that ej+1 2

is non negative it is sufficient to ensure that these three terms are non negative. This holds under the assumptions (7) and (8).

Indeed, using the definition of the fluxEjand owing to (3), we obtain

T0> ρj+1 2ej+12 1 − δ t δ xj+1 2 (γ − 1) [uj]−+ c(ej+1 2) √ 2 + [uj+1] ++c(ej+12) 2 !! − δ t δ xj+1 2 ρj+1 2ej+ 1 2 [λ+ (ej+1, uj+1)]++ [λ−(ej, uj)]−

where, due to (7), the rhs is non negative by virtue of (9) and (10) .

Next, we turn to Ti1. Using twice the Young inequality and bearing in mind the definition of ρj, we observe that

δ t 2 δ xj+i δ xj+1 2 Rj+i> δ xj+i 4δ xj+1 2 (uj+i− uj+i)2  ρj+i− δ t δ xj+i (F+ j+i+12−F − j+i−12)  . Hence, we have Ti1> δ xj+i 4δ xj+1 2

(uj+i− uj+i)2 ρj+i− δ t δ xj+i (F+ j+i+12−F − j+i−12) − δ t γ 2 √ 2 ρj+1 2c(ej+12) δ xj+i ! .

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6 Thierry Goudon, Julie Llobell and Sebastian Minjeaud

Coming back to (5), we write T1i>(uj+i− uj+i) 2 4δ xj+1

2



T2i,0+ T2i,1where, for k = 0, 1,

Ti,k2 = δ xj+i+k−1 2 2 ρj+i+k−12− δt F+ j+i+k−F − j+i+k−1 2 − δ t γ 2 √ 2ρj+i+k−12c(ej+12).

Note that a non negative term has been added to obtain a symmetric formulation in the above inequality. Due to (3) and (7) we get

F+ j+i+k−F − j+i+k−16 δ xj+i+k−1 2 γ δ t ρj+i+k−12,

and this allows us to write

Ti,k2 > δ xj+i+k−1 2 2γ ρj+i+k−12 γ − 1 − δ t δ xj+i+k−1 2 4 √ 2c(ej+12) ! .

We conclude by observing that this term is non negative by virtue of (8).

4 Numerical diffusion and energy conservation.

It is worth discussing the expression of the numerical diffusion produced by our scheme, see also the appendix in [1]. Let us set the following non negative quantity

Cj=        −uj if uj6 −c(ej) u2j+ c(ej)2 4c(ej) if |uj| < c(ej) uj if uj> c(ej) and the following notations for averaged quantities

{q}j= qj−1 2+ qj+12 2 and {q}j+12 = qj+ qj+1 2 .

DenotingF|.|=F+−F−, wich is a positive quantity, the mass and the momentum fluxes can be cast as the sum of a centered term and a diffusion term:

Fj= {ρ}juj− Cj 2  ρj+1 2− ρj−12  , Gj+1 2 = {F }j+12{u}j+12− {F|.|}j+1 2 2 uj+1− uj . Concerning the internal energy and kinetic energy fluxes, they become:

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Ej= {ρe}juj− Cj 2  ej+1 2ρj+12− ej−12ρj−12  , Hj+1 2 = {F }j+12{ u2 2 }j+12− {F|.|}j+1 2 2 u2j+1 2 − u2j 2 ! .

As a by-product, it is remarkable that the scheme properly deals with 1D-contact discontinuities: if the discrete velocity and pressure are constant in the neighborhood of xj+1

2, ie uj−1= uj= uj+1= uj+2= u and Πj−1/2= Πj+1/2= Πj+3/2= Π , then

the scheme guaranties that they remain constant in the neighborhood of this point at the next time, ie Πj+1/2= Π and uj+1= u = uj.

Let us now introduce the averaged energy in xj+1

2 and xjdefined by:

Ej+1 2 = ej+1 2 + δ xjρj u2j 2 + δ xj+1ρj+1 u2j+1 2 2δ xj+1 2ρj+12 and Ej= u2j 2 + δ xj+1 2ρj+ 1 2ej+ 1 2+ δ xj− 1 2ρj− 1 2ej− 1 2 2δ xjρj .

To obtain conservative equations for those quantities, we introduce the fluxes

Tj=Ej+ Hj+1 2+Hj−12 2 andT ∗ j+12 = Ej+1+Ej 2 +Hj+12− δ xj+1Rj+1− δ xjRj 4 .

Next we get the following consistent balance equations for ρjEjand ρj+1 2Ej+12: ρj+1 2Ej+12− ρj+12Ej+12 δ t + Tj+1−Tj δ xj+1 2 +uj+1{Π }j+1− uj{Π }j δ xj+1 2 = 0 and ρjEj− ρjEj δ t + T∗ j+12−T ∗ j−1 2 δ xj + Πj+1 2 {u}j+1 2− Πj− 1 2 {u}j−1 2 δ xj = 0.

5 Numerical simulations of Riemann problems.

We perform the numerical resolutions of some Riemann problems – see [4] – on the computational domain [0, 1]. The number of grid points is equal to 1000 and the time step is given by δ t = δ x/100. We take γ = 1.4. The initial data ρ, u, p are piecewise constant functions with a discontinuity located at x0= 0.5, according to the table below. In Figure 1, we represent the pressure pj+1

2

, velocity ujand internal energy ej+1

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8 Thierry Goudon, Julie Llobell and Sebastian Minjeaud

ρl ρr ul ur pl pr T Test #1 1 0.125 0 0 1 0.1 0.25 Test #2 1 1 0 0 1000 0.01 0.012 Test #3 5.99924 5.99242 19.5975 −6.19633 460.894 46.0950 0.035

Test #1, the so-called Sod test problem, is a mild test whose solution consists of a left rarefaction, a contact disconinuity and a right shock. Test #2 is a more severe test problem whose solution contains a left rarefaction, a contact discontinuity and a right shock. Test #3 corresponds to the collision of two strong shocks and consists of a left facing shock (travelling very slowly to the right), a right travelling contact discontinuity and a right travelling shock wave.

Test #1

Test #2

Test #3

Fig. 1 Numerical (solid lines) and exact (dotted lines) solutions: pressure (left), velocity (middle), internal energy (right).

References

1. Berthelin, F., Goudon, T., Minjeaud, S.: Kinetic schemes on staggered grids for barotropic Euler models: entropy-stability analysis. Math. Comput. 84, 2221–2262 (2015)

2. Herbin, R., Kheriji, W., Latch´e, J.C.: Staggered schemes for all speed flows. ESAIM: Proceed-ings. 35, 122–150 (2012)

3. Herbin, R., Latch´e, J.C., Nguyen, T.: Consistent explicit staggered schemes for compressible flows. Part II: The Euler equation. hal-00821069 (2013)

4. Toro, E.F.: Riemann solvers and numerical methods for fluid dynamics, third edn. Springer-Verlag, Berlin (2009)

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